package leetcodejava.top100likedquestions;

import org.junit.Test;
/**
 * This is the solution of No. xxx problem in the LeetCode,
 * the website of the problem is as follow:
 * https://leetcode-cn.com/problems/edit-distance/
 * <p>
 * The description of problem is as follow:
 * ==========================================================================================================
 * 给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。
 * <p>
 * 你可以对一个单词进行如下三种操作：
 * <p>
 * 插入一个字符
 * 删除一个字符
 * 替换一个字符
 * 示例 1:
 * <p>
 * 输入: word1 = "horse", word2 = "ros"
 * 输出: 3
 * 解释:
 * horse -> rorse (将 'h' 替换为 'r')
 * rorse -> rose (删除 'r')
 * rose -> ros (删除 'e')
 * 示例 2:
 * <p>
 * 输入: word1 = "intention", word2 = "execution"
 * 输出: 5
 * 解释:
 * intention -> inention (删除 't')
 * inention -> enention (将 'i' 替换为 'e')
 * enention -> exention (将 'n' 替换为 'x')
 * exention -> exection (将 'n' 替换为 'c')
 * exection -> execution (插入 'u')
 * <p>
 * 来源：力扣（LeetCode）
 * * ==========================================================================================================
 * @author zhangyu (zhangyuyu417@gmail.com)
 */
public class EditDistance72 {
    @Test
    public void testEditDistance() {
        String word1 = "horse";
        String word2 = "ros";
        int num = minDistance(word1, word2);
        System.out.println(num);
    }

    /**
     * 编辑距离最小步数
     *
     * @param word1 字符串1
     * @param word2 字符串2
     * @return 步数
     */
    public int minDistance(String word1, String word2) {
        int n = word1.length();
        int m = word2.length();
        if (n * m == 0) {
            return n + m;
        }
        int[][] d = new int[n + 1][m + 1];
        for (int i = 0; i < n + 1; i++) {
            d[i][0] = i;
        }
        for (int j = 0; j < m + 1; j++) {
            d[0][j] = j;
        }
        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < m + 1; j++) {
                int left = d[i - 1][j] + 1;
                int down = d[i][j - 1] + 1;
                int left_down = d[i - 1][j - 1];
                if (word1.charAt(i - 1) != word2.charAt(j - 1))
                    left_down += 1;
                d[i][j] = Math.min(left, Math.min(down, left_down));
            }
        }
        return d[n][m];
    }
}
